Two sides of a parallelogram have lengths of 6 and 4. Angle BAD is 69 deg.?

Question

Find the length of the diagonals AC and BD.
This question has me completly perplexed.

Drawing: http://img361.imageshack.us/img361/7348/pgramxc9.jpg

Thank you so much

Answer 1

From symmetry, you have all the angles
69 for two, and 180-69= 111 for the other two.
Use the law of cosines: for triangle abc with opposing angles A,B,C we have the length of the 3rd side c for given a,b from:

c = sqrt(a^2 +b^2 -2abcos(C))

In your case: a=4, b=6 and C = 69 so

BD = sqrt(36+16-48(cos(69)) = 5.9

For C = 111 we have:

AC = sqrt(36+16-48(cos(111))= 8.32

Answer 2

You need the cosine rule. Let BD = x, AC = y.
In triangle ABD: x^2 = 6^2 + 4^2 - 2*6*4*cos(69), so x = 5.899.
In triangle ADC, you know the angle at D is180-69 = 111. So y^2 = 6^2 + 4^2 - 2*6*4*cos(111), so y = 8.319.

Answer 3

if one is 69, then the other is 111

AC = sqrt[a^2 + b^2 - 2ab*cos(A)]
BD = sqrt[a^2 + b^2 - 2ab*cos(B)]

AC = sqrt(4^2 + 6^2 - 2(4 * 6)cos(111))
AC = sqrt(16 + 36 - 2(24)cos(111))
AC = sqrt(52 - 48cos(111))
AC = about 8 or about 8.31875

BD = sqrt(4^2 + 6^2 - 2(4 * 6)cos(69))
BD = sqrt(52 - 48cos(69))
BD = about 6 or about 5.89901

AC = about 8.319
BD = about 5.899

Info found at http://mathforum.org/dr.math/faq/formulas/faq.quad.html#parallelogram

Answer 4

bd^2=8^2+6^2-2*6*8cos 69=100-96cos69
ac^2=8^2+6^2-2-8-6cos 111=100-96cos111
bd=65.6
ac=134.4

Answer 5

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